Proving $R-S$ contains a prime ideal when $S$ is a multiplicative set I'm mainly trying to prove that 

If $0\not \in S\subseteq R$ is a multiplicative subset of a
  commutative ring $R$ with identity. Then $R-S$ contains a prime ideal.

Now, by using Zorn's lemma, one can show that $R-S$ contains a maximal ideal I. In fact, I is a prime ideal as well, so we need to show this.To do that, I assumed the opposite, i.e $\exists a,b \in R$ s.t $ab\in I$, but $a,b \not \in I$. Then looked each case. However, I stuck when $a,b \in R-[S\cup I]$. The following is what I've done for this case;
Proof:
In this case, our assumptions are $a,b\in R-[S\cup I]$ and $I + (a) = J$, where $J$ is an ideal in $R$ and $J \cap S \not = \emptyset$. 
These imply that; $S^{-1}(I + (a)) = S^{-1} J$, but since $J \cap S \not = \emptyset$, by a theorem that we proved in the class, $S^{-1} J = S^{-1}R$.Therefore, it must be true that $\exists (i\in I, r\in R, s_1,s \in S)$ s.t 
$$\frac{i+ra}{s_1} =\frac{1}{s}$$
but this implies $\exists s_2 \in S$ s.t 
$$ s_2(is + sra -1 s_1) = 0 \quad \Rightarrow \quad (s_1 s )i + (s_1 s r) a = (s_1 s )1,$$
,so
$$(b s_1 s )i + (s_1 s r) ab = (s_1 s )b.$$
Observe that by our assumption $ab\in I$, so $(s_1 s)b \in I$.
Now, at the point, I don't know how to proceed, so I would appreciate any help or hint because I stuck at this point for over a week now.
Edit
Thanks for the great answers; however, but I'm specifically interested in how to continue from the point where I stuck.
 A: Here's an answer that uses techniques that you might have not seen, but can be a motivation for learning them. 
Since $S$ is a multiplicative subset we can form the localization $S^{-1}R$, a nonzero ring together with a natural map
$$f: R \rightarrow S^{-1}R$$
such that $f(s)$ is a unit in $S^{-1}R$ for all $s \in S$.
Then $S^{-1}R$ contains a prime ideal $\mathfrak{p}$, and pulling back gives a prime ideal $\mathfrak{q}= f^{-1}(\mathfrak{p})$. If $s\in S$ is contained in $\mathfrak{q}$ then $f(s)$ would be a unit contained in $\mathfrak{p}$, which is a contradiction and shows that $\mathfrak{q}$ is the prime you want. 
A: If $\;I\;$ isn't a prime idea of $\;R\;$ , then there exist $\;a,b\in R\;$ s.t. $\;ab\in I\;$ but $\;a,b\notin I\;$ and from here we get
$$I\lneq \begin{cases}I+aR,\\I+bR\end{cases}\implies \exists\,i,j\in I\,,\,\,x,y,\in R\;\;s.t.\;\;i+ax\,,\,j+by\notin R\setminus S\implies$$
$$\begin{cases}i+ax=s_1\in S\\{}\\j+by=s_2\in S\end{cases}\implies S\ni s_1s_2=ij+iby+jax+abxy\in I\implies I\cap S\neq \emptyset$$
and this is a contradiction.
